When $$3^{333}$$ is divided by 11, the remainder is

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Accepted Answer

To find the remainder when $$3^{333}$$ is divided by 11, we use modular arithmetic and the concept of cyclicity.

### Step 1: Determine the cyclicity of $$3^n \mod 11$$  
We calculate the first few powers of 3 modulo 11:  
- $$3^1 \mod 11 = 3$$  
- $$3^2 \mod 11 = 9$$  
- $$3^3 \mod 11 = 27 \equiv 5 \mod 11$$  
- $$3^4 \mod 11 = 81 \equiv 4 \mod 11$$  
- $$3^5 \mod 11 = 243 \equiv 1 \mod 11$$

The powers of $$3^n$$ modulo 11 repeat every 5 terms, i.e., the cyclicity is 5.

### Step 2: Simplify the exponent modulo cyclicity  
Since the cyclicity is 5, we compute $$333 \mod 5$$:  
$$
333 \div 5 = 66 	ext{ remainder } 3
$$  
Thus, $$333 \equiv 3 \mod 5$$. This means $$3^{333} \equiv 3^3 \mod 11$$.

### Step 3: Calculate $$3^3 \mod 11$$  
From Step 1, we know that:  
$$
3^3 \equiv 27 \equiv 5 \mod 11
$$

### Final Answer:  
The remainder when $$3^{333}$$ is divided by 11 is **5**.  
**Correct Option:** D

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